The area enclosed by the curve $\frac{x^2}{16}+\frac{y^2}{25}=1$ is:

$20π$

$\frac{x^2}{16}+\frac{y^2}{25}=1$

as area is symmetric in all 4 quadrants

Area = 4 × area of 1st quadrant

$=4\sqrt{\frac{5^2}{4^2}}\int\limits_0^4\sqrt{4^2-x^2}dx$

$∵\frac{x^2}{16}+\frac{y^2}{25}=1⇒\frac{y^2}{5^2}=\frac{4^2-x^2}{5^2}$

$y^2=\frac{5^2}{4^2}(4^2-x^2)$

$y=\sqrt{\frac{5^2}{4^2}}\sqrt{4^2-x^2}$

$⇒4×\frac{5}{4}\left[(\frac{x}{2})\sqrt{4^2-x^2}+\frac{4^2}{2}\sin^{-1}(\frac{x}{4})\right]_0^4$

$4×\frac{5}{4}\left[\frac{4^2}{2}×\frac{π}{2}\right]=20π$ sq. units